Let $I$ be the incenter of a triangle $ABC$. The lines passing through $A$ and parallel to $BI, CI$ meet the perpendicular bisector to $AI$ at points $S, T$ respectively. Let $Y$ be the common point of $BT$ and $CS$, and $A^*$ be a point such that $BICA^*$ is a parallelogram. Prove that the midpoint of segment $YA^*$ lies on the excircle of the triangle touching the side $BC$.
Complex bash with unit circle $(ABC)$ works .
It is known that there exists $p, q, r$ such a $a=p^2, b=q^2 , c=r^2$ and $I$ has coordinate $-(pq+qr+rp)$, and also $b_1=-pr, c_1=-pq, j=pq+pr-qr$,there $J$ is $A-excenter$ and $B_1,C_1$ are middles of minor arcs $AC,AB$. I'm slightly lazy, so here are only formulas for coordinates.
We know that $S, T \in B_1C_1$, so we can find \[ s=p^2-pq-qr, t=p^2-pr-qr. \]For $Y$ we can find \[ y=-qr+(q+r) \frac{p^2(p-q-r)}{pq+pr-qr} \]and for $A^*$ we can get $a^*=q^2+r^2+pq+qr+rp$.
Now, \[m-j=\frac{(q+r)(p-q)(p-r)(p-q-r)}{2j} \]and \[ x-j=(q+r) (p-q) (p-r)/p.\]We need to prove that $(m-j)(\overline{m-j})=(x-j)(\overline{x-j}) $, which is now equivalent to $(p-q-r)(\overline{p-q-r})=j\bar{j}/p^2$, which is true since $|p|=|q|=|r|=1$ and problem is solved.